


Right rotation on a matrix is shifting each column to the right side (the last column moves to the first column) and 'K' times means performing this rotation 'K' times.
For 'K' = 1 and the given 'MAT':
1 2 3
4 5 6
7 8 9
Output after rotating 'MAT' one time is:
3 1 2
6 4 5
9 7 8
The first line of input contains an integer 'T' representing the number of test cases.
The first line of each test case contains three single space-separated integers ‘N’, ‘M’, and ‘K’, respectively. ‘N’ and ‘M’ represent the rows and columns of the matrix and ‘K’ denotes the number of right rotations to be performed.
Then each of the next 'N' lines of each test case contains 'M' single space-separated integers representing the elements in a row of the matrix.
For each test case, return the elements of the matrix row-wise after rotation in a single line.
You don't need to print the output, It has already been taken care of. Just implement the given function.
1 <= T <= 10
1 <= N <= 200
1 <= M <= 200
0 <= K <= 10^9
1 <= MAT[i][j] <= 10^5
Where 'MAT[i][j]' denotes the element in the 'i'th row and 'j'th column of the matrix.
It is guaranteed that sum of 'N' * 'M' over all the test cases does not exceed 10 ^ 5.
Time limit: 1 sec
This approach is based on the fact that when we rotate an array to the right by ‘K’ times, it shifts ‘K’ elements from the end to the beginning of the array while the remaining elements shift towards the end. The effective rotations in ‘MAT’ can be from 0 to 'M' - 1, as we get the same matrix ‘MAT’ after every 'M' rotations. So, we will set ‘K’ to ‘K’ % ‘M’.
Now, we traverse ‘MAT’ row-wise. We will follow the below steps to rotate the ‘i'-th row of ‘MAT’: